Question

A uniform hollow spherical ball of mass 17 kg and radius 50.0 cm rolls without slipping up a ramp at an angle of 19 degrees above the horizontal. The speed of the rolling ball at the base of the ramp is 5.0 m/s. While the ball is rolling up the ramp, what are the magnitude and direction of the acceleration of its center of mass, and what is the magnitude and direction of the friction force acting on the ball from the surface of the ramp

Answer 1

`a=-1.92 m/s^(2)`

`F_(f)=-21.76 N`

Step-by-step explanation:

We can use the definition of the torque:

`\tau=I\alpha`

When:

• I is the inertia of a uniform hallow sphere
  `I = (2/3)mR^(2)`
• α is the angular acceleration (α = a/R)

Now, torque is the product of the friction force times the radius.

`F_(f)*R=(2)/(3)mR^(2)*(a)/(R)`

`F_(f)=(2)/(3)ma` (1)

Now, let's analyze the force acting over the sphere using the Newton's second law.

`F=ma`

`-mgsin(\theta)-F_(f)=ma` (2)

Let's put F(f) of the equation (1) into the equation (2):

`-mgsin(\theta)-(2)/(3)ma=ma`

`a=-(3)/(5)gsin(\theta)`

`a=-(3)/(5)*9.81*sin(19)`

`a=-1.92 m/s^(2)`

Hence:
`F_(f)=(2)/(3)ma=(2)/(3)*17*(-1.92)`

`F_(f)=-21.76 N`

I hope it helps you!

Answer 2

The acceleration of its center of mass is

The frictional force is
`f = 21.65 \ N`

Step-by-step explanation:

From the question we are told that

The mass of the ball is
`m = 17 kg`

The radius of the ball is
`r = 50cm = (50)/(100) = 0.5m`

the angle with the horizontal is
`\theta = 19 ^ o`

The of the ball at the base is
`v = 5.0 \ m/s`

This setup is shown on the first uploaded image

looking at the diagram we see that the force acting on the ball can be mathematically evaluated as

`mg sin \theta -f = ma`

Where f is the frictional force

The torque on the ball is mathematically represented as

`\tau = f * r`

This torque can also be mathematically represented as

`\tau = I \alpha`

where I is the moment of inertia of the ball which is mathematically represented as

`I = (2)/(3) m r^2`

While
`\alpha` is the angular acceleration which is mathematically represented as

`\alpha = (a)/(r)`

So
`\tau = (2)/(3) m r^2 * (a)/(r)`

Equating the both formula for torque

`f * r = (2)/(3) m r^2 * (a)/(r )`

=>
`f = (2)/(3) ma`

Substituting this for f in the above equation

`mg sin \theta = ma + (2)/(3) ma`

`g sin \theta = (5)/(3) a`

`a = (3)/(5) * g * sin \theta \alpha`

Substituting values

`a = 1.91 m/s^2`

Now substituting into the equation frictional force equation

`f = (2)/(3) * 17 * 1.91`

`f = 21.65 \ N`